Question: $\dfrac{d}{dx}\left(4x-\dfrac{3}{x}+\dfrac{1}{x^2}\right)=$
The strategy We can first rewrite each rational term in the expression as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting rational terms as negative powers $\begin{aligned} &\phantom{=}4x-\dfrac{3}{x}+\dfrac{1}{x^2} \\\\ &=4x-3x^{-1}+x^{-2} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(4x-3x^{-1}+x^{-2}) \\\\ &=4\dfrac{d}{dx}(x)-3\dfrac{d}{dx}(x^{-1})+\dfrac{d}{dx}(x^{-2}) \\\\ &=4 (1x^0)-3(-1)x^{-2}+(-2)x^{-3} \\\\ &=4+3x^{-2}-2x^{-3} \\\\ &=4+\dfrac{3}{x^2}-\dfrac{2}{x^3} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(4x-\dfrac{3}{x}+\dfrac{1}{x^2}\right)=4+\dfrac{3}{x^2}-\dfrac{2}{x^3}$.